There is one special case
for factoring that you may or may not need, depending upon how your book
is structured and how your instructor intends to teach factoring quadratics.
I call it "factoring in pairs".

Factor xy – 5y – 2x + 10.

Is there anything that
factors out of all four terms? No. When you have four terms, and nothing
factors out of all of them, think of factoring "in pairs".
To factor "in pairs", I split the expression into two pairs
of terms, and then factor the pairs separately.

xy – 5y – 2x + 10

What can I factor out
of the first pair? I can take out a "y":

xy – 5y – 2x + 10

= y(x – 5) – 2x + 10

What can I factor out
of the second pair? I can take out a "–2":

xy – 5y – 2x + 10

= y(x – 5) – 2x + 10

= y(x – 5) – 2(x – 5)

(I took out a –2,
rather than a 2,
because the leading sign on the pair was a "minus". And I
got a "–5"
in the result because, when I divided the positive10 by the negative2,
the result was a negative5.
Be careful with your signs!)

Now that I do have a
common factor, I can proceed as usual:

xy – 5y – 2x + 10

= y(x – 5) – 2(x – 5)

=
(x – 5)(y – 2)

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Factoring "in pairs"
is most commonly used to introduce factoring quadratics. So you may see
exercises that look like this:

Factor x2 + 4x – x – 4.

This polynomial has four
terms with no factor common to all four, so I'll try to factor "in
pairs":

x2 + 4x – x – 4

= x(x + 4) – 1(x + 4)

= (x + 4)(x – 1)

In the second line above,
I factored a "1"
out. Why? If "nothing" factors out, a "1"
factors out.

You can use the Mathway widget below to
practice factoring a polynomial by grouping. Try the entered exercise,
or type in your own exercise. Then click the "paper-airplane" button to compare
your answer to Mathway's. (Or skip the widget and continue with the lesson.)

(Clicking on "Tap to view steps"
on the widget's answer screen will take you to the Mathway site for a paid upgrade.)

If you will be using "factoring in pairs" for factoring quadratics
(which is not the method I use), your book will refer to it as something like "factoring by
grouping", and it will work like this:

Factor x2 – 5x – 6.

First, I have to find
factors of the last term, –6, that add up to the numerical coefficient of the middle term, –5. I'll use the number –6 and +1,
because (–6)(+1)
= –6, and (–6)
+ (+1) = –5. Using
these numbers, I'll split the middle "–5x"
term into the two terms "–6x"
and "+1x".
This will then allow me to factor in pairs:

x2– 5x – 6

= x2– 6x + 1x – 6

= x(x – 6) + 1(x – 6)

= (x – 6)(x + 1)

Factor 6x2 –13x + 6.

This is a bit more complicated,
because the leading coefficient (the number on the x2) is
not a simple "1".
But I can still factor the polynomial.

First, I need to find
factors of (6)(6)
= 36 that add up
to –13.
I'll use the numbers –9 and –4,
because (–9)(–4)
= 36, and (–9) + (–4) = –13.
Then I can split the middle "–13x"
term into the two terms "–9x"
and "–4x",
and then factor in pairs:

6x2–13x + 6

= 6x2–9x – 4x + 6

= 3x(2x – 3) – 2(2x – 3)

= (2x – 3)(3x – 2)

For a complete explanation
of these last two examples (whose method may look somewhat "magical"
at the moment), please study my lesson on factoring
quadratics. The
"simple" and then the "hard" case pages should completely
clarify the topic.